Proof that every normed vector space is a topological vector space

The first point is fine. For the second, fix $(v_0,\alpha_0)\in V\times K$ and $\varepsilon >0$. We have to find $\delta>0$ such that if $|\alpha-\alpha_0|\leq \delta$ and $|v-v_0|\leq \delta$ then $\lVert \alpha_0v_0-\alpha v\rVert\leq \varepsilon$. We have \begin{align} \lVert \alpha_0v_0-\alpha v\rVert&\leq \lVert \alpha_0v_0-\alpha v_0\rVert+ \lVert \alpha v_0-\alpha v\rVert\\ &=|\alpha_0-\alpha|\lVert v_0\rVert+|\alpha|\lVert v-v_0\rVert\\ &\leq |\alpha_0-\alpha|(\lVert v_0\rVert+\lVert v-v_0\rVert)+|\alpha_0|\lVert v-v_0\rVert. \end{align} We take $\delta$ such that $\delta^2+\delta(\lVert v_0\rVert+|\alpha_0|)\leq \varepsilon$ (which is possible).

In this case, $\lVert \alpha_0v_0-\alpha v\rVert\leq \varepsilon$ when $|\alpha-\alpha_0|\leq \delta$ and $|v-v_0|\leq \delta$.